# Maximum of minimum for every window size

Posted: 21 Dec, 2020
Difficulty: Hard

## PROBLEM STATEMENT

#### For example:

``````ARR = [1,2,3,4]
Minimums of window size 1 = min(1), min(2), min(3), min(4) = 1,2,3,4
Maximum among (1,2,3,4)  is 4

Minimums of window size 2 = min(1,2), min(2,3),   min(3,4) = 1,2,3
Maximum among (1,2,3) is 3

Minimums of window size 3 = min(1,2,3), min(2,3,4) = 1,2
Maximum among (1,2) is 2

Minimums of window size 4 = min(1,2,3,4) = 1
Maximum among them is 1
The output array should be [4,3,2,1].
``````
##### Input Format:
``````The first line of the input contains an integer ‘T’ denoting the number of test cases.

The first line of each test case contains a single positive integer ‘N’ denoting the number of the elements present in the array.

The second line of each test case contains ‘N’ space-separated integers denoting the elements of the array.
``````
##### Output Format:
``````The only line of output of each test case should contain ‘N’ space-separated integer such that he ith integer indicates the maximum of minimums of the windows of size ‘i’.
``````
##### Constraints:
``````1 <= T <= 100
1 <= N <= 10 ^ 4
-10 ^ 9 <= ARR[i] <= 10 ^ 9

Where ‘T’ is the number of test cases, ‘N’ is the size of the array and ‘ARR[i]’ is the size of the array elements.

Time Limit: 1 sec
`````` Approach 1
• We will use two nested loops for sliding on the window of every possible size and one more inner loop to traverse on the window and store the minimum element of the current window in a ‘temp’ variable.
• We will create an array named ‘answer’. The ‘answer[i]’ will store the maximum of all the available minimum of every window size ‘i’.
• If ‘i’ and ‘j’  are the starting and ending indexes of the window then its length = j-i+1.
• So we update our ‘answer[length]’ with the maximum of all the available minimum of every window size ‘i’ with the help of a ‘temp’ variable